| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Vehicle on slope with resistance |
| Difficulty | Standard +0.3 This is a standard M1 mechanics problem requiring resolution of forces on a slope with an angled rope. Students must draw a force diagram, resolve parallel to the slope (including weight component, resistance, and tension component), and apply F=ma. While it involves multiple forces and angles, it follows a routine template taught extensively in M1 with no novel problem-solving required—slightly easier than average due to the 'show that' format which provides the target answer. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Diagram showing: Weight (\(mg\) or \(2000g\)) acting vertically downward | B1 | Must show arrow downward from particle |
| Normal reaction (\(R\)) perpendicular to slope, Tension (\(T\)) along rope at \(12°\) to slope, Resistance (500 N) along slope downward | B1 | All three remaining forces correct with labels |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Resolving along the slope: \(T\cos12° - 500 - 2000g\sin5° = 2000 \times 0.6\) | M1 | Equation of motion along slope, must include all relevant terms |
| Correct equation: \(T\cos12° = 2000 \times 0.6 + 500 + 2000 \times 9.8 \times \sin5°\) | A1 | Correct terms, allow \(g = 9.8\) or \(9.81\) |
| \(T\cos12° = 1200 + 500 + 1712.6...\) | A1 | Correct numerical evaluation of \(2000g\sin5°\) |
| \(T\cos12° = 3412.6...\) | A1 | Correct sum of terms on RHS |
| \(T = \dfrac{3412.6...}{\cos12°} = 3480\) N | A1 | \(T = 3480\) N correct to 3 significant figures, confirmed \(\checkmark\) |
## Question 8:
**Part (a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Diagram showing: Weight ($mg$ or $2000g$) acting vertically downward | B1 | Must show arrow downward from particle |
| Normal reaction ($R$) perpendicular to slope, Tension ($T$) along rope at $12°$ to slope, Resistance (500 N) along slope downward | B1 | All three remaining forces correct with labels |
---
**Part (b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Resolving along the slope: $T\cos12° - 500 - 2000g\sin5° = 2000 \times 0.6$ | M1 | Equation of motion along slope, must include all relevant terms |
| Correct equation: $T\cos12° = 2000 \times 0.6 + 500 + 2000 \times 9.8 \times \sin5°$ | A1 | Correct terms, allow $g = 9.8$ or $9.81$ |
| $T\cos12° = 1200 + 500 + 1712.6...$ | A1 | Correct numerical evaluation of $2000g\sin5°$ |
| $T\cos12° = 3412.6...$ | A1 | Correct sum of terms on RHS |
| $T = \dfrac{3412.6...}{\cos12°} = 3480$ N | A1 | $T = 3480$ N correct to 3 significant figures, confirmed $\checkmark$ |8 A van, of mass 2000 kg , is towed up a slope inclined at $5 ^ { \circ }$ to the horizontal. The tow rope is at an angle of $12 ^ { \circ }$ to the slope. The motion of the van is opposed by a resistance force of magnitude 500 newtons. The van is accelerating up the slope at $0.6 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{f30b02da-a41e-44cb-b45f-9e6a3a9d0528-22_269_991_513_529}
Model the van as a particle.
\begin{enumerate}[label=(\alph*)]
\item Draw a diagram to show the forces acting on the van.
\item Show that the tension in the tow rope is 3480 newtons, correct to three significant figures.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2011 Q8 [7]}}